How to get a path to a resource in a Java JAR file

Question

I am trying to get a path to a Resource but I have had no luck    This works  both in IDE and with the JAR  but this way I can t get a path to a file  only the file contents    ClassLoader classLoader   getClass   getClassLoader    PrintInputStream classLoader getResourceAsStream  config netclient p       If I do this   ClassLoader classLoader   getClass   getClassLoader    File file   new File classLoader getResource  config netclient p   getFile       The result is   java io FileNotFoundException  file  path to jarfile bot jar  config netclient p  No such file or directory   Is there a way to get a path to a resource file

User · Answer

I spent a while messing around with this problem  because no solution I found actually worked  strangely enough  The working directory is frequently not the directory of the JAR  especially if a JAR  or any program  for that matter  is run from the Start Menu under Windows  So here is what I did  and it works for  class files run from outside a JAR just as well as it works for a JAR   I only tested it under Windows 7    try         Attempt to get the path of the actual JAR file  because the working directory is frequently not where the file is        Example  file  D  all Java TitanWaterworks TitanWaterworks-en jar  TitanWaterworks class       Another example   D  all Java TitanWaterworks TitanWaterworks class     PROGRAM DIRECTORY   getClass   getClassLoader   getResource  TitanWaterworks class   getPath       Gets the path of the class or jar         Find the last   and cut it off at that location  If this isn t being run from a jar  there is no    so it ll cause an exception  which is fine      try           PROGRAM DIRECTORY   PROGRAM DIRECTORY substring 0  PROGRAM DIRECTORY lastIndexOf              catch  Exception e             Find the last   and cut it off at that location      PROGRAM DIRECTORY   PROGRAM DIRECTORY substring 0  PROGRAM DIRECTORY lastIndexOf        1         If it starts with    cut it off      if  PROGRAM DIRECTORY startsWith       PROGRAM DIRECTORY   PROGRAM DIRECTORY substring 1  PROGRAM DIRECTORY length           If it starts with file    cut that off  too      if  PROGRAM DIRECTORY startsWith  file      PROGRAM DIRECTORY   PROGRAM DIRECTORY substring 6  PROGRAM DIRECTORY length       catch  Exception e        PROGRAM DIRECTORY         Current working directory instead

User · Answer

When loading a resource make sure you notice the difference between  getClass   getClassLoader   getResource  quot com myorg foo jpg quot     relative path  and getClass   getResource  quot  com myorg foo jpg quot       note the slash at the beginning  I guess  this confusion is causing most of problems when loading a resource   Also  when you re loading an image it s easier to use getResourceAsStream    BufferedImage image   ImageIO read getClass   getResourceAsStream  quot  com myorg foo jpg quot      When you really have to load a  non-image  file from a JAR archive  you might try this  File file   null  String resource    quot  com myorg foo xml quot   URL res   getClass   getResource resource   if  res getProtocol   equals  quot jar quot          try           InputStream input   getClass   getResourceAsStream resource           file   File createTempFile  quot tempfile quot    quot  tmp quot            OutputStream out   new FileOutputStream file           int read          byte   bytes   new byte 1024            while   read   input read bytes      -1                out write bytes  0  read                     out close            file deleteOnExit          catch  IOException ex            Exceptions printStackTrace ex           else         this will probably work in your IDE  but not from a JAR     file   new File res getFile        if  file    null  amp  amp   file exists          throw new RuntimeException  quot Error  File  quot    file    quot  not found  quot

User · Answer

You need to understand the path within the jar file  Simply refer to it relative    So if you have a file  myfile txt   located in foo jar under the  src main resources directory  maven style    You would refer to it like   src main resources myfile txt   If you dump your jar using jar -tvf myjar jar  you will see the output and the relative path within the jar file  and use the FORWARD SLASHES

User · Answer

if netclient p is inside a JAR file  it won t have a path because that file is located inside other file  in that case  the best path you can have is really file  path to jarfile bot jar  config netclient p

User · Answer

Inside your ressources folder  java main resources  of your jar add your file  we assume that you have added an xml file named imports xml   after that you inject ResourceLoader if you use spring like bellow    Autowired private ResourceLoader resourceLoader    inside tour function write the bellow code in order to load file       Resource resource   resourceLoader getResource  classpath imports xml        try          File file          file   resource getFile     will load the file          catch IOException e  e printStackTrace

User · Answer

It may be a little late but you may use my library KResourceLoader to get a resource from your jar   File resource   getResource  file txt

User · Answer

A File is an abstraction for a file in a filesystem  and the filesystems don t know anything about what are the contents of a JAR   Try with an URI  I think there s a jar    protocol that might be useful for your purpouses

User · Answer

This is deliberate   The contents of the  file  may not be available as a file   Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource  The classloader does not have to provide a file handle to the resource  for example the jar file may not have been expanded into individual files in the file system   Anything you can do by getting a java io File could be done by copying the stream out into a temporary file and doing the same  if a java io File is absolutely necessary

User · Answer

This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names  File file   null  String resource    quot  view Trial main html quot    URL res   getClass   getResource resource   if  res toString   startsWith  quot jar  quot          try           InputStream input   getClass   getResourceAsStream resource           file   File createTempFile new Date   getTime    quot  quot    quot  html quot            OutputStream out   new FileOutputStream file           int read          byte   bytes   new byte 1024            while   read   input read bytes      -1                out write bytes  0  read                     out flush            out close            input close            file deleteOnExit          catch  IOException ex            ex printStackTrace            else         this will probably work in your IDE  but not from a JAR     file   new File res getFile

User · Answer

Maybe this method can be used for quick solution   public class TestUtility        public static File getInternalResource String relativePath                File resourceFile   null          URL location   TestUtility class getProtectionDomain   getCodeSource   getLocation            String codeLocation   location toString            try              if  codeLocation endsWith   jar                      Call from jar                 Path path   Paths get location toURI    resolve     classes     relativePath  normalize                    resourceFile   path toFile                 else                    Call from IDE                 resourceFile   new File TestUtility class getClassLoader   getResource relativePath  getPath                            catch URISyntaxException ex               ex printStackTrace                      return resourceFile

User · Answer

The following path worked for me  classpath  path to resource in jar

User · Answer

When in a jar file  the resource is located absolutely in the package hierarchy  not file system hierarchy   So if you have class com example Sweet loading a resource named    default conf  then the resource s name is specified as   com example default conf     But if it s in a jar then it s not a File

User · Answer

The one line answer is -   String path   this getClass   getClassLoader   getResource  lt resourceFileName gt   toExternalForm     Basically getResource method gives the URL   From this URL you can extract the path by calling toExternalForm    References   getResource     toExternalForm

User · Answer

private static final String FILE LOCATION    com input file somefile txt      Method Body   InputStream invalidCharacterInputStream   URLClassLoader getSystemResourceAsStream FILE LOCATION     Getting this from getSystemResourceAsStream is the best option   By getting the inputstream rather than the file or the URL  works in a JAR file and as stand alone

User · Answer

follow code   src main resources file streamToFile getClass   getClassLoader   getResourceAsStream  quot file quot     public static File streamToFile InputStream in        if  in    null            return null             try           File f   File createTempFile String valueOf in hashCode      quot  tmp quot            f deleteOnExit             FileOutputStream out   new FileOutputStream f           byte   buffer   new byte 1024            int bytesRead          while   bytesRead   in read buffer      -1                out write buffer  0  bytesRead                      return f        catch  IOException e            LOGGER error e getMessage    e           return null

User · Answer

In my case  I have used a URL object instead Path   File  File file   new File  my path    URL url   file toURI   toURL      Resource in classpath using classloader  URL url   MyClass class getClassLoader   getResource  resource name     When I need to read the content  I can use the following code   InputStream stream   url openStream      And you can access the content using an InputStream

[java] How to get a path to a resource in a Java JAR file

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